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Proof step 2 from proof reach-n-path-n (previous step)

⊦ reachable-n(H, n, r, r0) ↔ (∃ x. path(x, H) ∧ hd x = r ∧ x.last = r0 ∧ # x = n + 1), n = 0

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ H, r, r0. (reachable-n(H, n0, r, r0) ↔ (∃ x. path(x, H) ∧ hd x = r ∧ x.last = r0 ∧ # x = n0 + 1)) ∨ n0 = 0 ⊦ n0 +1 = 0, reachable-n(H, n0 +1, r, r0) ↔ (∃ x. path(x, H) ∧ hd x = r ∧ x.last = r0 ∧ # x = n0 +1 + 1)

Rule argument

indvar: n

Additional information

Used simplifier rules: nat freely generated by 0, +1;;